Delta fuction potential general solution

[Danny Boy]

Hi, in the book 'Introduction to Quantum Mechanics' by Griffiths, on page 71 in the section 'The Delta-Function Potential' he states that the general solution to time independent Schrodinger Equation is $$\psi(x) = Ae^{-\kappa x} + B e^{\kappa x}$$

he then notes that the first term blows up as $$x \to -\infty,$$ so we must choose $$A=0.$$ Why is it that we are rejecting a wave function that goes to infinity? Is it simply because we are looking for normalizable solutions?

Thanks.

 

[vanhees71]

For ##x<0## you must choose ##A=0## and for ##x>0## you must have ##B=0##, if ##\kappa>0##. That's indeed, because you want to have bound states in this case, leading to normalizable wave functions. Note that there may be solutions with ##\kappa \in \mathrm{i} \mathbb{R}##, leading to scattering states with energy eigenvalues in the continuum, and these wave functions are generalized eigenfunctions that are "normalizable to a ##\delta## distribution" only.

 

[Danny Boy]

For ##x<0## you must choose ##A=0## and for ##x>0## you must have ##B=0##, if ##\kappa>0##. That's indeed, because you want to have bound states in this case, leading to normalizable wave functions. Note that there may be solutions with ##\kappa \in \mathrm{i} \mathbb{R}##, leading to scattering states with energy eigenvalues in the continuum, and these wave functions are generalized eigenfunctions that are "normalizable to a ##\delta## distribution" only.

So my reasoning is correct then that we want normalizable solutions hence the requirement that the wave function is bounded?

 

[vanhees71]

If you restrict yourself to the bound-state solutions, it's correct.

 

[Danny Boy]

If you restrict yourself to the bound-state solutions, it's correct.

Okay thanks. One quick question, do you maybe know why the delta function is said to have units 1/length?

 

[Avalanche_]

Simple explanation: in electrodynamics when you want to write (for example) charge density function for point charge in the origin. You would write that as ρ [itex] = q \delta (x) \delta (y) \delta (z) [/itex] . You will write it in that way so that ∫ [itex] \rho dV[/itex] over the whole space gives you the result [itex] q[/itex]. But ρ must have units coulomb per volume, but in your formula you have coulombs in [itex] q [/itex], so delta functions have units 1/length